Polynomial maps with nilpotent Jacobians in dimension three II

Abstract

In the paper, we first classify all polynomial maps of the form $H=(u(x,y),v(x,y,z),h(x,y))$ in the case that $JH$ is nilpotent and $(\deg_{y}u,\deg_{y}h)\leq 3$, $H(0)=0$. Then we classify all polynomial maps of the form $H=(u(x,y,z),v(x,y,u),h(x,y))$ in the case that $JH$ is nilpotent and $(\deg v(x,y,0),\allowbreak\deg h)\leq 3$, $H(0)=0$. Finally, we classify polynomial maps of the form $H=(u(x,y,z),v(x,y,z),h(x,y))$ in certain conditions.

Keywords. Jacobian Conjecture, Nilpotent Jacobian matrix, Polynomial maps
MSC(2010). Primary 14E05; Secondary 14A05;14R15

1 Introduction

Throughout this paper, we will write ${\bf K}$ for algebraically closed field and ${\bf K}[X]={\bf K}[x_{1},x_{2},\ldots,x_{n}]$ (${\bf K}[\bar{X}]={\bf K}[x,y,z]$)for the polynomial algebra over ${\bf K}$ with $n$ ($3$) indeterminates. Let $F=(F_{1},F_{2},\ldots,F_{n}):{\bf{K}}^{n}\rightarrow{\bf{K}}^{n}$ be a polynomial map, that is, $F_{i}\in{\bf{K}}[X]$ for all $1\leq i\leq n$. Let $JF=(\frac{\partial F_{i}}{\partial x_{j}})_{n\times n}$ be the Jacobian matrix of $F$.

The Jacobian Conjecture (JC) raised by O.H. Keller in 1939 in [8] states that a polynomial map $F:{\bf{K}}^{n}\rightarrow{\bf{K}}^{n}$ is invertible if the Jacobian determinant $\det JF$ is a nonzero constant. This conjecture has been attacked by many people from various research fields, but it is still open, even for $n\geq 2$. Only the case $n=1$ is obvious. For more information about the wonderful 70-year history, see [1], [5], and the references therein.

In 1980, S.S.S.Wang ([9]) showed that the JC holds for all polynomial maps of degree 2 in all dimensions (up to an affine transformation). The most powerful result is the reduction to degree 3, due to H.Bass, E.Connell and D.Wright ([1]) in 1982 and A.Yagzhev ([11]) in 1980, which asserts that the JC is true if the JC holds for all polynomial maps $X+H$, where $H$ is homogeneous of degree 3. Thus, many authors study these maps and led to pose the following problem.

(Homogeneous) dependence problem. Let $H=(H_{1},\ldots,H_{n})\in{\bf K}[X]$ be a (homogeneous) polynomial map of degree $d$ such that $JH$ is nilpotent and $H(0)=0$. Whether $H_{1},\ldots,H_{n}$ are linearly dependent over ${\bf K}$?

The answer to the above problem is affirmative if rank$JH\leq 1$ ([1]). In particular, this implies that the dependence problem has an affirmative answer in the case $n=2$. D. Wright give an affirmative answer when $H$ is homogeneous of degree 3 in the case $n=3$ ([10]) and the case $n=4$ is solved by Hubbers in [7]. M. de Bondt and A. van den Essen give an affirmative answer to the above problem in the case $H$ is homogeneous and $n=3$ ([3]). A. van den Essen finds the first counterexample in dimension three for the dependence problem ([6]). M. de Bondt give a negative answer to the homogeneous dependence problem for $d\geq 3$. In particular, he constructed counterexamples to the problem for all dimensions $n\geq 5$ ([2]). In [4], M. Chamberland and A. van den Essen classify all polynomial maps of the form $H=(u(x,y),v(x,y,z),h(u(x,y),v(x,y,z)))$ with $JH$ nilpotent. In particular, they show that all maps of this form with $H(0)=0$, $JH$ nilpotent and $H_{1},H_{2},H_{3}$ are linearly independent has the same form as the counterexample that gave by A. van den Essen in [6] (up to a linear coordinate change). We classify all polynomial maps of the form $H=(u(x,y),v(x,y,z),h(x,y,z))$ in the case that $JH$ is nilpotent and $\deg_{z}v\leq 3$, $(\deg_{y}u(x,y),\deg_{y}h(x,y,z))=1$ ([12]) and classify all polynomial maps of the form $H=(u(x,y),v(x,y,z),h(x,y))$ in the case that $JH$ is nilpotent and $(\deg_{y}u(x,y),\deg_{y}h(x,y))\leq 2$ or $\deg_{y}u(x,y)$ or $\deg_{y}h(x,y)$ is a prime number ([13]).

In section 2, we classify all polynomial maps of the form $H=(u(x,y),v(x,y,z),\allowbreak h(x,y))$ in the case that $JH$ is nilpotent and $(\deg_{y}u,\deg_{y}h)\leq 3$, $H(0)=0$. Then, in section 3, we classify all polynomial maps of the form $H=(u(x,y,z),v(x,y,u),\allowbreak h(x,y))$ in the case that $JH$ is nilpotent and $(\deg v(x,y,0),\deg h)\leq 3$, $H(0)=0$. We prove that $u,v,h$ are linearly dependent in the case that $JH$ is nilpotent and $H$ has the form: $H=(u(x,y,z),v(x,y,z),h(x,y))$ with $\deg_{z}v=1$ and $\deg_{z}u\geq 2$, $H(0)=0$ in section 4. The main results in the paper are Theorem 2.2, Theorem 2.5, Theorem 3.2 and Theorem 3.3, Theorem 4.1. We define $Q_{x_{i}}:=\frac{\partial Q}{\partial x_{i}}$ and that $\deg_{y}f$ is the highest degree of $y$ in $f$.

2 Polynomial maps of the form $H=(u(x,y),v(x,\allowbreak y,z),h(x,y))$

In this section, we classify all polynomial maps of the form $H=(u(x,y),v(x,y,z),\allowbreak h(x,y))$ in the case that $JH$ is nilpotent and $(\deg_{y}u,\deg_{y}h)\leq 3$, $H(0)=0$.

Lemma 2.1.

Let $Q\in{\bf K}[x,y]$ and $Q(x,y)=Q_{r}(x)y^{r}+Q_{r-1}(x)y^{r-1}+\cdots+Q_{1}(x)y+Q_{0}(x)$ with $Q_{r}(x)\in{\bf K}^{*}$. If $Q_{y}|Q_{x}$ or $Q_{y}|Q+c$ for some $c\in{\bf K}$, then $Q$ is a polynomial of $y+a(x)$ for some $a(x)\in{\bf K}[x]$.

Proof.

Let $Q_{r}:=Q_{r}(x)$. Then we have $Q_{y}=rQ_{r}y^{r-1}+(r-1)Q_{r-1}(x)y^{r-2}+\cdots+Q_{1}(x)$ and $Q_{x}=Q_{r-1}^{\prime}(x)y^{r-1}+\cdots+Q_{1}^{\prime}(x)y+Q_{0}^{\prime}(x)$. Thus, we have $\deg_{y}Q_{y}\geq\deg_{y}Q_{x}$. We always view that the polynomials are in ${\bf K}[x][y]$ with coefficients in ${\bf K}[x]$ when comparing the coefficients of $y^{i}$.

Case I If $Q_{y}|Q_{x}$, then we have

$$Q_{x}=b(x)Q_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.1)$$

for some $b(x)\in{\bf K}[x]$. That is,

$$Q_{r-1}^{\prime}(x)y^{r-1}+\cdots+Q_{1}^{\prime}(x)y+Q_{0}^{\prime}(x)=b(x)(rQ_{r}y^{r-1}+\cdots+Q_{1}(x))$$

Comparing the coefficients of $y^{r-1}$ of the above equation, we have that $b(x)=\frac{Q_{r-1}^{\prime}(x)}{rQ_{r}}$. Then let $\bar{x}=x$, $\bar{y}=y+\frac{Q_{r-1}(x)}{rQ_{r}}$, it follows from equation $(2.1)$ that $Q_{\bar{x}}=0$. That is, $Q(x,y)\in{\bf K}[y+\frac{Q_{r-1}(x)}{rQ_{r}}]$. Let $a(x)=\frac{Q_{r-1}(x)}{rQ_{r}}$. Then the conclusion follows.

Case II If $Q_{y}|(Q+c)$, then $Q+c=(e(x)y+d(x))Q_{y}$. That is,
$Q_{r}y^{r}+Q_{r-1}(x)y^{r-1}+\cdots+Q_{1}(x)y+Q_{0}(x)+c=(e(x)y+d(x))(rQ_{r}y^{r-1}+(r-1)Q_{r-1}(x)y^{r-2}+\cdots+Q_{1}(x))$                         (2.2)
Comparing the coefficients of $y^{r}$ and $y^{r-1}$ of equation $(2.2)$, we have that $e(x)=\frac{1}{r}$ and $d(x)=\frac{Q_{r-1}(x)}{r^{2}Q_{r}}$. Then equation $(2.2)$ has the following form:
$Q_{r}y^{r}+Q_{r-1}(x)y^{r-1}+\cdots+Q_{1}(x)y+Q_{0}(x)+c=(\frac{1}{r}y+\frac{Q_{r-1}(x)}{r^{2}Q_{r}})(rQ_{r}y^{r-1}+(r-1)Q_{r-1}(x)y^{r-2}+\cdots+Q_{1}(x))$                         (2.3)

Claim: $Q_{i}(x)=C_{r}^{i}\frac{Q_{r-1}^{r-i}(x)}{r^{r-i}Q_{r}^{r-i-1}(x)}$ for $1\leq i\leq r-2$ and $C_{r}^{i}=\frac{r!}{i!(r-i)!}$.

Comparing the coefficients of $y^{r-2}$ of equation $(2.3)$, we have the following equation:

$$Q_{r-2}(x)=C_{r}^{r-2}\frac{Q_{r-1}^{2}(x)}{r^{2}Q_{r}}$$

Suppose $Q_{k+1}(x)=C_{r}^{k+1}\frac{Q_{r-1}^{r-k-1}(x)}{r^{r-k-1}Q_{r}^{r-k-2}(x)}$. Then comparing the coefficients of $y^{k}$ of equation $(2.3)$, we have the following equation:

$$Q_{k}(x)=\frac{1}{r}kQ_{k}(x)+\frac{Q_{r-1}(x)}{r^{2}Q_{r}}(k+1)Q_{k+1}(x).$$

That is,

$$Q_{k}(x)=C_{r}^{k}\frac{Q_{r-1}^{r}(x)}{r^{r-k}Q_{r}^{r-k-1}}.$$

Thus, we have that $Q_{i}(x)=C_{r}^{i}\frac{Q_{r-1}^{r-i}(x)}{r^{r-i}Q_{r}^{r-i-1}(x)}$ for $1\leq i\leq r-2$. Then equation $(2.3)$ has the following form: $Q_{0}(x)+c=Q_{1}(x)\frac{Q_{r-1}(x)}{r^{2}Q_{r}}$. That is,

$$Q_{0}(x)=\frac{Q_{r-1}^{r}(x)}{r^{r}Q_{r}^{r-1}}-c.$$

Thus, we have $Q(x,y)=Q_{r}(y^{r}+\frac{Q_{r-1}(x)}{Q_{r}}y^{r-1}+\cdots+\frac{Q_{r-1}^{r}(x)}{r^{r}Q_{r}^{r}})-c=Q_{r}(y+\frac{Q_{r-1}(x)}{rQ_{r}})^{r}-c$. Therefore, we have $Q\in{\bf K}[y+\frac{Q_{r-1}(x)}{rQ_{r}}]$. Let $a(x)=\frac{Q_{r-1}(x)}{rQ_{r}}$. Then the conclusion follows. ∎

Theorem 2.2.

Let $H=(u(x,y),v(x,y,z),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and $(\deg_{y}u,\deg_{y}h)\leq 3$, then $u=g(ay+b(x))$, $v=v_{1}z-a^{-1}b^{\prime}(x)g(ay+b(x))-v_{1}l_{2}x$, $h=c_{0}u^{2}+l_{2}u$, where $b(x)=v_{1}c_{0}ax^{2}+l_{1}x+\tilde{l}_{2}$; $v_{1},c_{0},a\in{\bf K}^{*}$; $l_{1},l_{2},\tilde{l}_{2}\in{\bf K}$, $g(t)\in{\bf K}[t]$ and $g(0)=0$, $\deg_{t}g(t)\geq 1$.

Proof.

If $(\deg_{y}u,\deg_{y}h)\leq 2$, then the conclusion follows from Theorem 3.5 in [13]. Let $v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}$. Then it follows from Lemma 3.2 in [13] that $d=1$ and $v_{1}\in{\bf K}^{*}$. Since $JH$ is nilpotent, so we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.$$

It follows from equation $(2.4)$ that $u_{x}=-v_{0y}$. Thus, there exists $P\in{\bf K}[x,y]$ such that

$$u=-P_{y},~{}~{}v=P_{x}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.7)$$

It follows from equation $(2.6)$ and Lemma 3.1 in [13] that there exists $q\in{\bf K}[x,y]$ such that

$$u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)$$

Since $\deg_{y}q|(\deg_{y}u,\deg_{y}h)$ and $(\deg_{y}u,\deg_{y}h)=3$, so we have $\deg_{y}q=1$ or 3.

If $\deg_{y}q=1$, then the conclusion follows from the proof of Theorem 2.8 in [12].

If $\deg_{y}q=3$, then $\deg_{y}q_{y}=2$. Let $q(x,y)=q_{3}(x)y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)$.
Claim: $q_{3}(x)\in{\bf K}^{*}$.
Let $P(x,y)=a_{r}(x)y^{r}+a_{r-1}(x)y^{r-1}+\cdots+a_{1}(x)y+a_{0}(x)$. It follows from Lemma 3.4 in [13] that $a_{r}(x)\in{\bf K}^{*}$. It follows from equations $(2.7)$ and $(2.8)$ that $u(x,y)=u(q)=-P_{y}=-(ra_{r}(x)y^{r-1}+\cdots+a_{1}(x))$. Thus, we have $q_{3}(x)\in{\bf K}^{*}$. Substituting equations $(2.7)$ and $(2.8)$ to equation $(2.5)$, we have the following equation:

$$q_{y}[v_{1}h^{\prime}(q)+v_{0x}u^{\prime}(q)]=-(u^{\prime}(q)q_{x})^{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.9)$$

(1) If $q_{y}$ is irreducible, then we have $q_{y}|q_{x}$ or $q_{y}|u^{\prime}(q)$. Since $u^{\prime}(q)$ is a polynomial of $q$, so we have $u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k})$ with $c_{0}\in{\bf K}^{*}$ and $c_{i}\in{\bf K}$, $1\leq i\leq k$. Since $q_{y}$ is irreducible, so there exists $i_{0}\in\{1,2,\ldots,k\}$ such that $q_{y}|q+c_{i_{0}}$ in the case that $q_{y}|u^{\prime}(q)$. It follows from Lemma 2.1 that $q$ is a polynomial of $y+a(x)$ for some $a(x)\in{\bf K}[x]$. That is, $u,h\in{\bf K}[y+a(x)]$. Then the conclusion follows from the proof of Theorem 2.8 in [12].

(2) If $q_{y}$ is reducible, then

$$q_{y}=3q_{3}y^{2}+2q_{2}(x)y+q_{1}(x)=3q_{3}(y+b(x))(y+e(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.10)$$

for some $b(x),~{}e(x)\in{\bf K}[x]$. Clearly, $y+b(x)$, $y+e(x)$ are irreducible. It follows from equation $(2.9)$ that we have the following cases:
Case I $q_{y}|q_{x}$
Case II $q_{y}|u^{\prime}(q)$
Case III $(y+b(x))|q_{x}$ and $(y+e(x))|u^{\prime}(q)$
Case IV $(y+b(x))|q_{x}$ in the case $b(x)=e(x)$.
Case V $(y+b(x))|u^{\prime}(q)$ in the case $b(x)=e(x)$.

Case I If $q_{y}|q_{x}$, then the conclusion follows from Lemma 2.1.

Case II If $q_{y}|u^{\prime}(q)$, then $(y+b(x))(y+e(x))|u^{\prime}(q)$. Since $u^{\prime}(q)$ is a polynomial of $q$, so $u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k})$ for $c_{0}\in{\bf K}^{*}$ and $c_{i}\in{\bf K}$, $1\leq i\leq k$. Thus, there exist $i_{0},~{}j_{0}\in\{1,2,\ldots,k\}$ such that $(y+b(x))|q+c_{i_{0}}$ and $(y+e(x))|q+c_{j_{0}}$. That is,

$$q_{3}y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)+c_{i_{0}}=(y+b(x))(q_{3}y^{2}+d_{1}(x)y+d_{0}(x))~{}~{}~{}~{}~{}(2.11)$$

and

$$q_{3}y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)+c_{j_{0}}=(y+e(x))(q_{3}y^{2}+f_{1}(x)y+f_{0}(x))~{}~{}~{}~{}~{}(2.12)$$

We always view that the polynomials are in ${\bf K}[x][y]$ with coefficients in ${\bf K}[x]$ in the following arguments. Comparing the coefficients of $y^{2}$, $y$ of equations $(2.11)$ and $(2.12)$, we have the following equations:

$$q_{2}(x)=d_{1}(x)+q_{3}b(x)=f_{1}(x)+q_{3}e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.13)$$

and

$$q_{1}(x)=d_{0}(x)+d_{1}(x)b(x)=f_{0}(x)+e(x)f_{1}(x)~{}~{}~{}~{}~{}(2.14)$$

It follows from equations $(2.13)$ and $(2.14)$ that

$$d_{0}(x)=q_{1}(x)-b(x)q_{2}(x)+q_{3}b^{2}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.15)$$

and

$$f_{0}(x)=q_{1}(x)-e(x)q_{2}(x)+q_{3}e^{2}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.16)$$

Then equations $(2.11)$ and $(2.12)$ have the following forms:

$$q_{0}(x)+c_{i_{0}}=b(x)d_{0}(x)=b(x)(q_{1}(x)-b(x)q_{2}(x)+q_{3}b^{2}(x))~{}~{}~{}~{}~{}~{}~{}~{}(2.17)$$

and

$$q_{0}(x)+c_{j_{0}}=e(x)f_{0}(x)=e(x)(q_{1}(x)-e(x)q_{2}(x)+q_{3}e^{2}(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.18)$$

Then equation $(2.17)-(2.18)$ has the following form:

$$c_{i_{0}}-c_{j_{0}}=(b(x)-e(x))[q_{1}(x)-q_{2}(x)(b(x)+e(x))+q_{3}(b^{2}(x)+b(x)e(x)+e^{2}(x))].$$

Thus, we have $b(x)-e(x)\in{\bf K}$. That is, $e(x)=b(x)+c$ for some $c\in{\bf K}$. Therefore, we have $q_{y}=3q_{3}(y+b(x))(y+b(x)+c)=3q_{3}(y+b(x))^{2}+3q_{3}c(y+b(x))$. Then $q(x,y)=q_{3}(y+b(x))^{3}+\frac{3}{2}q_{3}c(y+b(x))^{2}+f(x)$ for some $f(x)\in{\bf K}[x]$. Since $(y+b(x))|(q+c_{i_{0}})$, so we have $(y+b(x))|(f(x)+c_{i_{0}})$. That is, $f(x)=-c_{i_{0}}$. Thus, $q$ is a polynomial of $y+b(x)$. Therefore, $u,h\in{\bf K}[y+b(x)]$. Then the conclusion follows from the proof of Theorem 2.8 in [12]

Case III Since $u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k})$ for $c_{0}\in{\bf K}^{*}$ and $c_{i}\in{\bf K}$, $1\leq i\leq k$. Clearly, $y+e(x)$ is irreducible, so there exists $i_{0}\in\{1,2,\ldots,k\}$ such that $(y+e(x))|(q+c_{i_{0}})$. That is,

$$q^{\prime}_{2}(x)y^{2}+q_{1}^{\prime}(x)y+q^{\prime}_{0}(x)=(y+b(x))(q^{\prime}_{2}(x)y+d(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.19)$$

and

$$q_{3}y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)+c_{i_{0}}=(y+e(x))(q_{3}y^{2}+w_{1}(x)y+w_{0}(x))~{}~{}~{}~{}~{}~{}(2.20)$$

where $d(x),w_{1}(x),w_{0}(x)\in{\bf K}[x]$. Comparing the coefficients of $y$, $y^{0}$ of equation $(2.19)$, we have the following equations:

$$q^{\prime}_{1}(x)-b(x)q^{\prime}_{2}(x)=d(x)$$

and

$$q^{\prime}_{0}(x)=b(x)d(x)=b(x)(q^{\prime}_{1}(x)-b(x)q^{\prime}_{2}(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.21)$$

Comparing the coefficients of $y$, $y^{0}$ of equation $(2.10)$, we have the following equations:

$$\frac{2q_{2}(x)}{3q_{3}}=b(x)+e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.22)$$

and

$$\frac{q_{1}(x)}{3q_{3}}=b(x)e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.23)$$

Substituting equations $(2.22)$ and $(2.23)$ to equation $(2.21)$, we have the following equation:

$$q^{\prime}_{0}(x)=q_{3}(3b(x)b^{\prime}(x)e(x)+\frac{3}{2}b^{2}(x)e^{\prime}(x)-\frac{3}{2}b^{2}(x)b^{\prime}(x))$$

Integrating the two sides of the above equation with respect to $x$, we have the following equation:

$$q_{0}(x)=q_{3}(\frac{3}{2}b^{2}(x)e(x)-\frac{1}{2}b^{3}(x))+\bar{c}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.24)$$

for some $\bar{c}\in{\bf K}$. Comparing the coefficients of $y^{2}$, $y$ and $y^{0}$ of equation $(2.20)$, we have the following equations:

$$\left\{\begin{aligned} q_{2}(x)=w_{1}(x)+q_{3}e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.25)\\ q_{1}(x)=w_{0}(x)+w_{1}(x)e(x)~{}~{}~{}~{}~{}~{}~{}(2.26)\\ q_{0}(x)+c_{i_{0}}=e(x)w_{0}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.27)\\ \end{aligned}\right.$$

It follows from equations $(2.25)$ and $(2.26)$ that

$$w_{0}(x)=q_{1}(x)-e(x)q_{2}(x)+q_{3}e^{2}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.28)$$

Substituting equations $(2.28)$, $(2.22)$ and $(2.23)$ to equation $(2.27)$, we have the following equation:

$$q_{0}(x)=q_{3}(\frac{3}{2}b(x)e^{2}(x)-\frac{1}{2}e^{3}(x))-c_{i_{0}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.29)$$

Comparing equation $(2.29)$ with equation $(2.24)$, we have $\frac{1}{2}q_{3}[b^{3}(x)-3b^{2}(x)e(x)+3b(x)e^{2}(x)-e^{3}(x)]=\bar{c}+c_{i_{0}}$. That is,

$$\frac{1}{2}q_{3}(b(x)-e(x))^{3}=\bar{c}+c_{i_{0}}.$$

Thus, we have $b(x)-e(x)\in{\bf K}$. That is, $b(x)=e(x)+\bar{\bar{c}}$ for some $\bar{\bar{c}}\in{\bf K}$. It follows from equation $(2.10)$ that $q_{y}=3q_{3}(y+e(x))^{2}+3q_{3}\bar{\bar{c}}(y+e(x))$. Then we have $q=q_{3}(y+e(x))^{3}+\frac{3}{2}q_{3}\bar{\bar{c}}(y+e(x))^{2}+\bar{f}(x)$ for some $\bar{f}(x)\in{\bf K}[x]$. Since $(y+e(x))|(q+c_{i_{0}})$, so we have $(y+e(x))|(\bar{f}(x)+c_{i_{0}})$. Thus, we have $\bar{f}(x)=-c_{i_{0}}$. Therefore, $q$ is a polynomial of $y+e(x)$. That is, $u,h,q\in{\bf K}[y+e(x)]$. Then the conclusion follows from the proof of Theorem 2.8 in [12].

Case IV, V If $b(x)=e(x)$, then it follows from equation $(2.10)$ that $q_{y}=3q_{3}(y+b(x))^{2}$. Thus, we have $q=q_{3}(y+b(x))^{3}+\bar{\bar{f}}(x)$ for some $\bar{\bar{f}}(x)\in{\bf K}[x]$.

If $(y+b(x))|q_{x}$, then $(y+b(x))|(3q_{3}b^{\prime}(x)(y+b(x))^{2}+\bar{\bar{f}}^{\prime}(x))$. Thus, we have $\bar{\bar{f}}^{\prime}(x)=0$. That is, $\bar{\bar{f}}(x)\in{\bf K}$. Therefore, we have $q\in{\bf K}[y+b(x)]$.

If $(y+b(x))|u^{\prime}(q)$, then it follows from the arguments of Case II that $(y+b(x))|q+c_{i_{0}}$ for some $c_{i_{0}}\in{\bf K}$. That is, $(y+b(x))|(q_{3}(y+b(x))^{3}+\bar{\bar{f}}(x)+c_{i_{0}})$. Thus, we have $\bar{\bar{f}}(x)=-c_{i_{0}}\in{\bf K}$. Therefore, $q\in{\bf K}[y+b(x)]$.
Thus, $u,~{}h$ are polynomials of $y+b(x)$ in the two cases. Then the conclusion follows from the proof of Theorem 2.8 in [12]. ∎

Remark 2.3.

We can replace the condition that $(\deg_{y}u(x,y),m)=1$ by the condition $(\deg_{y}u(x,y),m)\leq 3$ in Theorem 2.10 and replace the condition that $(m,n)=1$ by the condition $(m,n)\leq 3$ in Theorem 3.2 and Theorem 3.4 in [12].

Corollary 2.4.

Let $H=(u(x,y),v(x,y,z),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and $\deg_{y}u\leq 7$ or $\deg_{y}u=9$ or $\deg_{y}h\leq 7$ or $\deg_{y}h=9$, then $H$ has the form of Theorem 2.2.

Proof.

Let $v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}$. Then it follows from Lemma 3.2 in [13] that $d=1$ and $v_{1}\in{\bf K}^{*}$. Since $JH$ is nilpotent, so we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.$$

It follows from equation $(2.6)$ and Lemma 3.1 in [13] that there exists $q\in{\bf K}[x,y]$ such that

$$u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)$$

Thus, we have $\deg_{y}q|(\deg_{y}u,\deg_{y}h)$.

If $\deg_{y}u\leq 4$ or $\deg_{y}h\leq 4$, then the conclusion follows from Corollary 3.9 in [13].

If $\deg_{y}u=5$ or $\deg_{y}u=7$ or $\deg_{y}h=5$ or $\deg_{y}h=7$, then the conclusion follows from Theorem 3.3 in [13].

If $\deg_{y}u=6$ or 9 or $\deg_{y}h=6$ or 9, then $\deg_{y}q=1$ or 2 or 3 or 6 or 9.

Case I If $\deg_{y}q=1$ or 2 or 3, then the conclusion follows from the proof Theorem 2.2.

Case II If $\deg_{y}q=6$, then $\deg_{y}u=6$ or $\deg_{y}h=6$.

(1) If $\deg_{y}q=\deg_{y}u$, then it follows from equation $(2.8)$ that $u(x,y)=u(q)=\lambda q+\lambda_{0}$ with $\lambda\in{\bf K}^{*}$, $\lambda_{0}\in{\bf K}$. That is, $q=\lambda^{-1}u-\lambda^{-1}\lambda_{0}$. Thus, $h$ is a polynomial of $u$. Then the conclusion follows from Theorem 2.1 in [4].

(2) If $\deg_{y}q=\deg_{y}h$, then it follows from the arguments of (1) that $u$ is a polynomial of $h$. It follows from Corollary 2.3 in [13] that $u,v,h$ are linearly dependent. This is a contradiction!

Case III If $\deg_{y}q=9$, then $\deg_{y}u=9$ or $\deg_{y}h=9$. That is, $\deg_{y}q=\deg_{y}u$ or $\deg_{y}q=\deg_{y}h$. Then the conclusion follows from the arguments of Case II. ∎

Theorem 2.5.

Let $H=(u(x,y),v(x,y,z),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and the variety $V(Q_{y})$ is irreducible for any $Q_{y}|(u_{y},h_{y})$ and $\deg_{y}Q|(\deg_{y}u,\deg_{y}h)$, then $H$ has the form of Theorem 2.2.

Proof.

Let $v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}$. Then it follows from Lemma 3.2 in [13] that $d=1$ and $v_{1}\in{\bf K}^{*}$. Since $JH$ is nilpotent, so we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.$$

It follows from equation $(2.4)$ that $u_{x}=-v_{0y}$. Thus, there exists $P\in{\bf K}[x,y]$ such that

$$u=-P_{y},~{}~{}v=P_{x}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.7)$$

It follows from equation $(2.6)$ and Lemma 3.1 in [13] that there exists $q\in{\bf K}[x,y]$ such that

$$u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)$$

If $\deg_{y}q\leq 1$, then the conclusion follows from the proof of Theorem 2.8 in [12].

Suppose $\deg_{y}q\geq 2$. Let $q(x,y)=q_{t}(x)y^{t}+q_{t-1}(x)y^{t-1}+\cdots+q_{1}(x)y+q_{0}(x)$ with $t\geq 2$. It follows from equation $(2.8)$ and Lemma 3.4 in [13] that

$$q_{t}(x)\in{\bf K}^{*}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.30)$$

It follows from equations $(2.5)$, $(2.7)$ and $(2.8)$ that

$$q_{y}[v_{1}h^{\prime}(q)+v_{0x}u^{\prime}(q)]=-(u^{\prime}(q)q_{x})^{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.31)$$

Since $u_{y}=u^{\prime}(q)q_{y}$, $h_{y}=h^{\prime}(q)q_{y}$, so we have $q_{y}|(u_{y},h_{y})$ and $\deg_{y}q|(\deg_{y}u,\deg_{y}h)$. Thus, the variety $V(q_{y})$ is irreducible. That is, there exists an irreducible polynomial $f(x,y)\in{\bf K}[x,y]$ such that

$$q_{y}=f^{s}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.32)$$

for some $s\in{\bf N}^{*}$. It follows from equation $(2.31)$ that $q_{y}|(u^{\prime}(q))^{2}q^{2}_{x}$. That is, $f^{s}(x,y)|(u^{\prime}(q))^{2}q^{2}_{x}$. Since $f(x,y)$ is irreducible, so we have $f(x,y)|u^{\prime}(q)$ or $f(x,y)|q_{x}$.

Case I Suppose $f(x,y)|u^{\prime}(q)$. Since $u^{\prime}(q)$ is a polynomial of $q$, so it follows from the Fundamental Theorem of Algebra that $u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k})$ for $c_{0}\in{\bf K}^{*}$ and $c_{i}\in{\bf K}$, $1\leq i\leq k$. If $f(x,y)|u^{\prime}(q)$, then there exists $i_{0}\in\{1,2,\ldots,k\}$ such that $f(x,y)|(q+c_{i_{0}})$. That is,

$$q+c_{i_{0}}=f(x,y)M^{(1)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.33)$$

for some $M^{(1)}(x,y)\in{\bf K}[x,y]$. Thus, it follows from equations $(2.33)$ and $(2.32)$ that $q_{y}=f_{y}M^{(1)}+fM^{(1)}_{y}=f^{s}$. That is,

$$f_{y}M^{(1)}=f(f^{s-1}-M^{(1)}_{y}).$$

Since $f$ is irreducible and $f_{y}\neq 0$, so it follows from the above equation that $f|M^{(1)}$. That is,

$$M^{(1)}(x,y)=f(x,y)M^{(2)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.34)$$

for some $M^{(2)}(x,y)\in{\bf K}[x,y]$. Thus, it follows from equations $(2.33)$ and $(2.34)$ that

$$q+c_{i_{0}}=f^{2}(x,y)M^{(2)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.35)$$

Then it follows from equation $(2.35)$ that $q_{y}=2ff_{y}M^{(2)}+f^{2}M^{(2)}_{y}=f^{s}$. That is,

$$2f_{y}M^{(2)}=f(f^{s-2}-M^{(2)}_{y}).$$

Since $f$ is irreducible and $f_{y}\neq 0$, so we have $f|M^{(2)}$. We can do step by step until

$$q+c_{i_{0}}=f^{s+1}M^{(s+1)}$$

for some $M^{(s+1)}\in{\bf K}[x,y]$. Then it follows from the above equation and equation $(2.32)$ that $q_{y}=(s+1)f^{s}f_{y}M^{(s+1)}+f^{s+1}M^{(s+1)}_{y}=f^{s}$. That is,

$$(s+1)f_{y}M^{(s+1)}+fM^{(s+1)}_{y}=1~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.36)$$

Let $f(x,y)=f_{l}(x)y^{l}+f_{l-1}(x)y^{l-1}+\cdots+f_{1}(x)y+f_{0}(x)$, $M^{(s+1)}=M^{(s+1)}_{m}(x)y^{m}+M^{(s+1)}_{m-1}(x)y^{m-1}+\cdots+M^{(s+1)}_{1}(x)y+M^{(s+1)}_{0}(x)$ with $f_{l}(x)M^{(s+1)}_{m}(x)\neq 0$ and $l\geq 1$. It follows from equations $(2.32)$ and $(2.30)$ that $f^{s}_{l}(x)=tq_{t}(x)\in{\bf K}^{*}$. That is, $f_{l}(x)\in{\bf K}^{*}$. It follows from equation $(2.36)$ that $(s+1)(lf_{l}y^{l-1}+(l-1)f_{l-1}(x)y^{l-2}+\cdots+f_{1}(x))(M^{(s+1)}_{m}(x)y^{m}+M^{(s+1)}_{m-1}(x)y^{m-1}+\cdots+M^{(s+1)}_{1}(x)y+M^{(s+1)}_{0}(x))+(f_{l}(x)y^{l}+f_{l-1}(x)y^{l-1}+\cdots+f_{1}(x)y+f_{0}(x))(mM^{(s+1)}_{m}(x)y^{m-1}+(m-1)M^{(s+1)}_{m-1}(x)y^{m-2}+\cdots+M^{(s+1)}_{1}(x))=1$                          (2.37)
We always view that the polynomials are in ${\bf K}[x][y]$ with coefficients in ${\bf K}[x]$ when comparing the coefficients of $y^{j}$.

Suppose $m\geq 1$. Comparing the coefficients of $y^{l+m-1}$ of equation $(2.37)$, we have the following equation:

$$(s+1)lf_{l}M^{(s+1)}_{m}(x)+mf_{l}M^{(s+1)}_{m}(x)=0$$

Since $f_{l}(x)M^{(s+1)}_{m}(x)\neq 0$, so we have $(s+1)l+m=0$. This is a contradiction because $sl\geq 1$ and $m\geq 1$. Therefore, we have $m=0$. That is, $M^{(s+1)}_{y}=0$. Then equation $(2.36)$ has the following form:

$$(s+1)f_{y}M^{(s+1)}=1$$

Thus, we have $f_{y}\in{\bf K}^{*}$. That is, $f=f_{1}y+f_{0}(x)$ with $f_{1}\in{\bf K}^{*}$. So $f$ is a polynomial of $y+a(x)$ with $a(x)=f_{1}^{-1}f_{0}(x)$. That is, $q,~{}u,~{}h\in{\bf K}[y+a(x)]$. Then the conclusion follows from the proof of Theorem 2.8 in [12].

Case II If $f(x,y)|q_{x}$, then

$$q_{x}=f(x,y)N^{(1)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.38)$$

for some $N^{(1)}(x,y)\in{\bf K}[x,y]$. Thus, it follows from equations $(2.38)$ and $(2.32)$ that $q_{xy}=f_{y}N^{(1)}+fN^{(1)}_{y}$, $q_{yx}=sf^{s-1}f_{x}$. Thus, we have the following equation:

$$f_{y}N^{(1)}=f(sf^{s-2}f_{x}-N^{(1)}_{y}).$$

Since $f$ is irreducible and $f_{y}\neq 0$, so we have $f|N^{(1)}$. That is, $N^{(1)}=fN^{(2)}$ for some $N^{(2)}\in{\bf K}[x,y]$. It follows from equation $(2.38)$ that $q_{x}=f^{2}N^{(2)}(x,y)$. We can do step by step until $q_{x}=f^{s}N^{(s)}(x,y)$. Then $q_{xy}=sf^{s-1}f_{y}N^{(s)}+f^{s}N^{(s)}_{y}=q_{yx}=sf^{s-1}f_{x}$. That is,

$$s(f_{x}-f_{y}N^{(s)})=fN^{(s)}_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.39)$$

Since $q_{y}=f^{s}$, $q_{x}=f^{s}N^{(s)}(x,y)$ and $\deg_{y}q_{x}\leq t-1=\deg_{y}q_{y}$, so we have $\deg_{y}(f^{s}N^{(s)}(x,y))\leq\deg_{y}(f^{s})$. Thus, we have $\deg_{y}N^{(s)}(x,y)=0$. That is, $N^{(s)}_{y}=0$. Then equation $(2.39)$ has the following form:

$$f_{x}=f_{y}N^{(s)}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.40)$$

where $N^{(s)}(x):=N^{(s)}(x,0)=N^{(s)}(x,y)$. Let $\bar{x}=x$, $\bar{y}=y+\int N^{(s)}(x)dx$. Then it follows from equation $(2.40)$ that $f_{\bar{x}}=0$. That is, $f\in{\bf K}[y+a(x)]$, where $a(x)=\int N^{(s)}(x)dx\in{\bf K}[x]$. Thus, $q,~{}u,~{}h$ are polynomials of $y+a(x)$. Then the conclusion follows from the proof of Theorem 2.8 in [12]. ∎

Corollary 2.6.

Let $H=(u(x,y),v(x,y,z),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and the variety $V(u_{y})$ or the variety $V(h_{y})$ is irreducible, then $H$ has the form of Theorem 2.2.

Proof.

Let $v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}$. Then it follows from Lemma 3.2 in [13] that $d=1$ and $v_{1}\in{\bf K}^{*}$. Since $JH$ is nilpotent, so we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.$$

It follows from equation $(2.6)$ and Lemma 3.1 in [13] that there exists $q\in{\bf K}[x,y]$ such that

$$u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)$$

Since $V(u_{y})$ or $V(h_{y})$ is irreducible and $V(q_{y})\subset V(u_{y})\cap V(h_{y})$ is a subvariety, so $V(q_{y})$ is irreducible. Then the conclusion follows from the proof of Theorem 2.5. ∎

3 Polynomial maps of the form $H=(u(x,y,z),v(x,\allowbreak y,u),h(x,y))$

In this section, we classify polynomial maps of the form $H=(u(x,y,z),v(x,y,u),\allowbreak h(x,y))$ in the case that $JH$ is nilpotent and $(\deg v(x,y,0),\deg h)\leq 3$. Combining Theorem 2.2 with Corollary 3.7 in [13], we have the following Proposition.

Proposition 3.1.

Let $H=(u(x,y,z),v(x,y),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and $(\deg_{x}v,\deg_{x}h)\leq 3$ or at least one of $\deg_{x}v$, $\deg_{x}h$ is a prime, then $u=u_{1}z-a^{-1}b^{\prime}(y)g(ax+b(y))-u_{1}l_{2}y$, $v=g(ax+b(y))$, $h=c_{0}v^{2}+l_{2}v$, where $b(y)=u_{1}c_{0}ay^{2}+l_{1}y+\tilde{l}_{2}$, $u_{1},c_{0},a\in{\bf K}^{*}$, $l_{1},l_{2},\tilde{l}_{2}\in{\bf K}$, $g(t)\in{\bf K}[t]$ and $g(0)=0$, $\deg_{t}g(t)\geq 1$.

In the following theorem, we denote $v_{x}=\frac{\partial v}{\partial x}(x,y,u)$, $v_{y}=\frac{\partial v}{\partial y}(x,y,u)$ and $v_{u}=\frac{\partial v}{\partial u}(x,y,u)$.

Theorem 3.2.

Let $H=(u(x,y,z),v(x,y,u),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and $\deg_{u}v\geq 2$. If $JH$ is nilpotent, then $u,v,h$ are linearly dependent.

Proof.

If $u_{z}=0$, then it follows from Proposition 2.1 in [12] that $u,v,h$ are linearly dependent. Thus, we can assume that $u_{z}\neq 0$. Since $JH$ is nilpotent, we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{y}=-v_{u}u_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.1)\\ u_{x}v_{y}-v_{x}u_{y}-h_{x}u_{z}=h_{y}v_{u}u_{z}~{}~{}~{}~{}~{}~{}~{}~{}(3.2)\\ u_{z}(v_{x}h_{y}-v_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.3)\end{aligned}\right.$$

Let $u=u_{d}z^{d}+u_{d-1}z^{d-1}+\cdots+u_{1}z+u_{0}$, $v(x,y,u)=v_{m}(x,y)u^{m}+v_{m-1}(x,y)u^{m-1}+\cdots+v_{1}(x,y)u+v_{0}(x,y)$ with $u_{d}v_{m}\neq 0$ and $d\geq 1$, $m\geq 2$, $u_{i},v_{j}\in{\bf K}[x,y]$ for $0\leq i\leq d$, $0\leq j\leq m$. It follows from equation $(3.1)$ that $u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{my}u^{m}+v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y}=-(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})$              (3.4)
We always view that the polynomials are in ${\bf K}[x,y][z]$ with coefficients in ${\bf K}[x,y]$ in the following arguments.

Since $m\geq 2$, comparing the coefficients of $z^{md}$ of equation $(3.4)$, we have $v_{my}u_{d}+mv_{m}u_{dy}=0$. That is,

$$\frac{v_{my}}{v_{m}}=-m\frac{u_{dy}}{u_{d}}$$

Suppose $v_{my}\neq 0$. Then $u_{dy}\neq 0$. Thus, we have $v_{m}u^{m}_{d}=e^{c(x)}$ by integrating the two sides of the above equation with respect to $y$, where $c(x)$ is a function of $x$. Since $v_{m}~{},u_{d}\in{\bf K}[x,y]$ and $e^{c(x)}$ is a function of $x$. Thus, we have $v_{m}~{},u_{d}\in{\bf K}[x]$. This is a contradiction! Therefore, we have $v_{my}=0=u_{dy}$. Then we have $u_{iy}=0$ by comparing the coefficients of $z^{d(m-1)+i}$ of equation $(3.4)$ for $i=d-1,~{}d-2,\ldots,1$. Then equation $(3.4)$ has the following form:
$u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y}=-(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})u_{0y}$                                        (3.5)
Comparing the coefficients of $z^{jd}$ for $j=m-1,~{}m-2,\ldots,2$ of equation $(3.5)$, we have the following equations:

$$v_{jy}+(j+1)v_{j+1}u_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.6)$$

Then we have

$$(v_{1y}+2v_{2}u_{0y})u_{k}=-u_{kx}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.7)$$

by comparing the coefficients of $z^{k}$ of equation $(3.5)$ for $k=d,~{}d-1,\ldots,1$. If $v_{1y}+2v_{2}u_{0y}=0$, then $u_{kx}=0$ for $1\leq k\leq d$. If $v_{1y}+2v_{2}u_{0y}\neq 0$, then we have $u_{kx}=0$ by comparing the degree of $x$ of two sides of equation $(3.7)$ for $1\leq k\leq d$. Since $u_{d}\neq 0$, we have $v_{1y}+2v_{2}u_{0y}=0$. This is a contradiction! Thus, we have $v_{1y}+2v_{2}u_{0y}=0$ and $u_{dx}=\cdots=u_{1x}=0$. Therefore, we have $u_{d},\ldots,u_{1}\in{\bf K}^{*}$. Then equation $(3.5)$ has the following form:

$$u_{0x}+v_{0y}+v_{1}u_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.8)$$

Since $u_{z}\neq 0$, so it follows from equation (3.3) that $h_{y}v_{x}-h_{x}v_{y}=0$. That is,

$$h_{y}(v_{mx}u^{m}+\cdots+v_{1x}u+v_{0x})-h_{x}(v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y})=0~{}~{}~{}~{}~{}(3.9)$$

Then we have $h_{y}v_{mx}=0$ by comparing the coefficients of $z^{md}$ of equation $(3.9)$. Thus, we have $h_{y}=0$ or $v_{mx}=0$.

(1) If $h_{y}=0$, then it follows from equation $(3.9)$ that $h_{x}=0$ or $v_{(m-1)y}=\cdots=v_{1y}=v_{0y}=0$.

(i) If $h_{x}=0$, then $h=0$ because $h(0)=0$. Thus, $u,v,h$ are linearly dependent.

(ii) If $v_{(m-1)y}=\cdots=v_{0y}=0$, then $v_{y}=0$. It follows from equation $(3.2)$ that $u_{y}v_{x}+h_{x}u_{z}=0$. That is, $u_{0y}(v_{mx}u^{m}+v_{(m-1)x}u^{m-1}+\cdots+v_{1x}u+v_{0x})+h_{x}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})=0$                      (3.10)

If $u_{0y}=0$, then it follows from equation $(3.10)$ that $h_{x}=0$. Since $H(0)=0$, we have $h=0$. Thus, $u,v,h$ are linearly dependent.

If $u_{0y}\neq 0$, then we have $v_{mx}=\cdots=v_{1x}=0$ by comparing the coefficients of $z^{md},z^{(m-1)d},\ldots,z^{d}$ of equation $(3.10)$ respectively.
If $d\geq 2$, then we have $h_{x}=0$ by comparing the coefficient of $z^{d-1}$ of equation $(3.10)$. Since $H(0)=0$, we have $h=0$. Thus, $u,v,h$ are linearly dependent.
If $d=1$, then it follows from equation $(3.6)$ ($j=m-1$) that $u_{0y}=0$. This is a contradiction!

(2) If $v_{mx}=0$, then $v_{m}\in{\bf K}^{*}$. It follows from equation $(3.2)$ that
$u_{0x}(v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y})-u_{0y}(v_{(m-1)x}u^{m-1}+\cdots+v_{1x}u+v_{0x})-h_{x}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})-h_{y}(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})=0$                       (3.11)

If $d\geq 2$, then we have $h_{y}=0$ by comparing the coefficient of $z^{d(m-1)+d-1}$ of equation $(3.11)$. Thus, we have $u_{0x}v_{ly}-u_{0y}v_{lx}=0$ by comparing the coefficients of $z^{ld}$ of equation $(3.11)$ for $l=m-1,m-2,\ldots,1$. Comparing the coefficients of $z^{d-1}$ of equation $(3.11)$, we have $h_{x}=0$. Thus, we have $h=0$ because $h(0)=0$. So $u,v,h$ are linearly dependent.

If $d=1$, then equation $(3.11)$ has the following form:
$u_{0x}(v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y})-u_{0y}(v_{(m-1)x}u^{m-1}+\cdots+v_{1x}u+v_{0x})-h_{x}u_{1}-h_{y}u_{1}(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})=0$                                  (3.12)
Thus, we have

$$u_{0x}v_{ly}-u_{0y}v_{lx}-h_{y}u_{1}(l+1)v_{l+1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.13)$$

by comparing the coefficients of $z^{ld}$ of equation $(3.12)$ for $l=m-1,m-2,\ldots,1$. Then equation $(3.12)$ has the following form:

$$u_{0x}v_{0y}-u_{0y}v_{0x}-u_{1}(h_{x}+v_{1}h_{y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.14)$$

Comparing the coefficients of $z^{ld}$ of equation $(3.9)$ for $l=m-1,m-2,\ldots,1$, we have

$$h_{y}v_{lx}-h_{x}v_{ly}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.15)$$

for $l=m-1,\ldots,1$. Then equation $(3.9)$ has the following form:

$$h_{y}v_{0x}-h_{x}v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.16)$$

It follows from equation $(3.6)$ that

$$v_{m-1}=-mv_{m}u_{0}+c_{1}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.17)$$

for some $c_{1}(x)\in{\bf K}[x]$. Substituting equation (3.17) to equation (3.13) for $l=m-1$, we have

$$u_{0y}c^{\prime}_{1}(x)+h_{y}mv_{m}u_{1}=0$$

Thus, we have $u_{0}c^{\prime}_{1}(x)+mv_{m}u_{1}\cdot h+e(x)=0$ by integrating the two sides of the above equation with respect to $y$, where $e(x)\in{\bf K}[x]$. That is,

$$h=-\frac{1}{mv_{m}u_{1}}c^{\prime}_{1}(x)u_{0}-\frac{1}{mv_{m}u_{1}}e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.18)$$

Substituting equations $(3.17)$ and $(3.18)$ to equation $(3.15)$ for $l=m-1$, we have

$$c^{\prime\prime}_{1}(x)u_{0}u_{0y}+\frac{1}{mv_{m}}(c^{\prime}_{1}(x))^{2}u_{0y}+e^{\prime}(x)u_{0y}=0$$

If $u_{0y}=0$, then it follows from equation $(3.18)$ that $h_{y}=0$. Then it reduces to (1).
If $u_{0y}\neq 0$, then we have

$$c^{\prime\prime}_{1}(x)u_{0}+\frac{1}{mv_{m}}(c^{\prime}_{1}(x))^{2}+e^{\prime}(x)=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.19)$$

Thus, we have $c_{1}^{\prime\prime}(x)=0$ and

$$e^{\prime}(x)=-\frac{1}{mv_{m}}(c_{1}^{\prime}(x))^{2}\in{\bf K}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.20)$$

by comparing the degree of $y$ of the equation (3.19). Then we have

$$c_{1}^{\prime}(x)(u_{0x}v_{0y}-u_{0y}v_{0x})+e^{\prime}(x)v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.21)$$

by substituting equation $(3.18)$ to equation $(3.16)$. Substituting equation (3.20) to equation (3.21), we have

$$c_{1}^{\prime}(x)(u_{0x}v_{0y}-u_{0y}v_{0x})-\frac{1}{mv_{m}}(c_{1}^{\prime}(x))^{2}v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.22)$$

If $c_{1}^{\prime}(x)=0$, then it follows from equation (3.20) that $e(x)\in{\bf K}$. Thus, it follows from equation (3.18) that $h=-\frac{e(x)}{mv_{m}u_{1}}\in{\bf K}$. Since $h(0,0)=0$, we have $h=0$. Therefore, $u,v,h$ are linearly dependent.

If $c_{1}^{\prime}(x)\neq 0$, then equation (3.22) has the following form:

$$u_{0x}v_{0y}-u_{0y}v_{0x}=\frac{1}{mv_{m}}c_{1}^{\prime}(x)v_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.23)$$

Substituting equations (3.18) and (3.23) to equation (3.14), we have the following equation:

$$c_{1}^{\prime}(x)(v_{0y}+u_{0x}+v_{1}u_{0y})+e^{\prime}(x)=0.$$

Substituting equation (3.8) to the above equation, we have $e^{\prime}(x)=0$. It follows from equation (3.20) that $c_{1}^{\prime}(x)=0$. This is a contradiction! ∎

Theorem 3.3.

Let $H=(u(x,y,z),v(x,y,u),h(x,y))$ be a polynomial map over ${\bf K}[x,y,z]$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and $(\deg v(x,y,0),\deg h)\leq 3$, then there exists $T\in\operatorname{GL}_{3}({\bf K})$ such that $T^{-1}HT$ has the form of Proposition 3.1.

Proof.

It follows from Theorem 3.2 that $\deg_{u}v\leq 1$. If $u_{z}=0$ or $v_{u}=0$, then it reduces to Proposition 2.1 in [12] and Proposition 3.1 respectively. Thus, we can assume that $u_{z}\neq 0$ and $v_{u}\neq 0$. Since $JH$ is nilpotent, we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{y}=-v_{u}u_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.1)\\ u_{x}v_{y}-v_{x}u_{y}-h_{x}u_{z}=h_{y}v_{u}u_{z}~{}~{}~{}~{}~{}~{}~{}~{}(3.2)\\ u_{z}(v_{x}h_{y}-v_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.3)\\ \end{aligned}\right.$$

Let $u=u_{d}z^{d}+u_{d-1}z^{d-1}+\cdots+u_{1}z+u_{0}$ and $v=v_{1}u+v_{0}$ with $u_{d}v_{1}\neq 0$.

It follows from equation (3.1) that $u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{1y}u+v_{0y}+v_{1}(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})=0$               (3.24)
Then we have

$$u_{dx}+v_{1y}u_{d}+v_{1}u_{dy}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.25)$$

and

$$u_{(d-1)x}+v_{1y}u_{d-1}+v_{1}u_{(d-1)y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.26)$$

by comparing the coefficients of $z^{d}$ and $z^{d-1}$ of equation (3.24) respectively.

It follows from equation (3.2) that
$(u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x})(v_{1y}u+v_{0y})-(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})(v_{1x}u+v_{0x})-h_{x}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})-h_{y}v_{1}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})=0$                                   (3.27)
Comparing the coefficients of $z^{k}$ of equation (3.27) for $k=2d,2d-1,\ldots,d+1$, we have

$$u_{(k-d)x}v_{1y}-u_{(k-d)y}v_{1x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.28)$$

Then we have

$$(u_{0x}v_{1y}-u_{0y}v_{1x})u_{d}+u_{dx}v_{0y}-u_{dy}v_{0x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.29)$$

and

$$(u_{0x}v_{1y}-u_{0y}v_{1x})u_{d-1}+u_{(d-1)x}v_{0y}-u_{(d-1)y}v_{0x}-du_{d}(h_{x}+v_{1}h_{y})=0~{}~{}~{}~{}~{}~{}(3.30)$$

by comparing the coefficients of $z^{d}$ and $z^{d-1}$ of equation (3.27) respectively.

It follows from equation (3.3) that

$$h_{y}(v_{1x}u+v_{0x})-h_{x}(v_{1y}u+v_{0y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.31)$$

Then we have

$$h_{y}v_{1x}-h_{x}v_{1y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.32)$$

and

$$h_{y}v_{0x}-h_{x}v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.33)$$

by comparing the coefficients of $z^{d}$ and $z^{0}$ of equation (3.31) orderly.

If $h_{y}\neq 0$, then let $\lambda(x,y)=\frac{h_{x}}{h_{y}}$. Thus, we have $h_{x}=\lambda(x,y)h_{y}$. It follows from equations (3.32) and (3.33) that

$$v_{1x}=\lambda(x,y)v_{1y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.34)$$

and

$$v_{0x}=\lambda(x,y)v_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.35)$$

It follows from equations (3.28) and (3.34) that

$$(u_{(k-d)x}-\lambda(x,y)u_{(k-d)y})v_{1y}=0$$

If $v_{1y}=0$, then it follows from equation (3.32) that $v_{1x}=0$ because $h_{y}\neq 0$. That is, $v_{1}\in{\bf K}^{*}$.

If $v_{1y}\neq 0$, then we have

$$u_{(k-d)x}=\lambda(x,y)u_{(k-d)y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.36)$$

for $d+1\leq k\leq 2d$. Substituting equations (3.34), (3.35), (3.36)$(k=2d)$ to equation (3.29), we have the following equation:

$$u_{0x}=\lambda(x,y)u_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.37)$$

Substituting equations (3.34), (3.35), (3.36), (3.37) to equation (3.30), we have the following equation:

$$h_{x}+v_{1}h_{y}=0.$$

That is,

$$v_{1}=-\frac{h_{x}}{h_{y}}=-\lambda(x,y).$$

Substituting equation (3.34) to the above equation, we have

$$v_{1}v_{1y}=-v_{1x}.$$

Then we have $v_{1}\in{\bf K}^{*}$ by comparing the degree of $x$ of two sides of the above equation. This is a contradiction!
Consequently, we have $H=(u,v_{1}u+v_{0}(x,y),h(x,y))$ with $v_{1}\in{\bf K}^{*}$. Let

$$T=\left(\begin{array}[]{ccc}1&0&0\\ v_{1}&1&0\\ 0&0&1\\ \end{array}\right).$$

Then $T^{-1}HT=(u(x,y+v_{1}x,z),v_{0}(x,y+v_{1}x),h(x,y+v_{1}x))$ and $\deg_{x}v_{0}(x,y+v_{1}x)=\deg v(x,y,0)$, $\deg_{x}h(x,y+v_{1}x)=\deg h(x,y)$. Since $(\deg v(x,y,0),\deg h)\leq 3$, so the conclusion follows from Proposition 3.1.

If $h_{y}=0$, then it follows from equation (3.32) that $h_{x}=0$ or $v_{1y}=0$.

(1) If $h_{x}=0$, then $h=0$ because $h(0,0)=0$. Thus, $u,v,h$ are linearly dependent. This is a contradiction!

(2) If $v_{1y}=0$, then it follows from equation $(3.28)$ that $v_{1x}=0$ or $u_{dy}=u_{(d-1)y}=\cdots=u_{1y}=0$.

Case I If $v_{1x}=0$, then $v_{1}\in{\bf K}^{*}$. Thus, the conclusion follows from the former arguments.

Case II If $u_{dy}=u_{(d-1)y}=\cdots=u_{1y}=0$, then it follows from equations (3.25) and (3.26) that $u_{dx}=0=u_{(d-1)x}$. Thus, it follows from equation (3.29) that $v_{1x}=0$ or $u_{0y}=0$. If $v_{1x}=0$, then we have $v_{1}\in{\bf K}^{*}$, it reduces to Case I. If $u_{0y}=0$, then it follows from equation (3.30) that $h_{x}=0$. Thus, we have $h=0$ because $h(0,0)=0$. Therefore, $u,v,h$ are linearly dependent. This is a contradiction! ∎

Corollary 3.4.

Let $H=(u(x,y,v),v(x,y,z),h(x,y))$ be a polynomial map with $H(0)=0$. Assume that the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent and $(\deg u(x,y,0),\deg h(x,y))\leq 3$, then there exists $T\in\operatorname{GL}_{3}({\bf K})$ such that $T^{-1}HT$ has the form of Proposition 3.1.

Proof.

Let

$$T_{1}=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\\ \end{array}\right).$$

Then $T_{1}^{-1}HT_{1}=(v(y,x,z),u(y,x,v(y,x,z)),h(y,x))$. Since $JH$ is nilpotent, we have that $J(T_{1}^{-1}HT_{1})=T_{1}^{-1}JHT_{1}$ is nilpotent. Since $\deg u(y,x,0)=\deg u(x,y,0)$, $\deg h(y,x))=\deg h(x,y)$ and $(\deg u(x,y,0),\deg h(x,y))\leq 3$, so it follows from Theorem 3.3 that there exists $T_{2}\in\operatorname{GL}_{3}({\bf K})$ such that $T_{2}^{-1}(T_{1}^{-1}HT_{1})T_{2}$ is of the form of Proposition 3.1. Let $T=T_{1}T_{2}\in\operatorname{GL}_{3}({\bf K})$. Then the conclusion follows. ∎

4 Polynomial maps of the form $H=(u(x,y,z),v(x,\allowbreak y,z),h(x,y))$

In the section, we classify polynomial maps of the form $H=(u(x,y,z),v(x,y,z),\allowbreak h(x,y))$ in the case that $JH$ is nilpotent and $\deg_{z}v(x,y,z)\leq 1$. Firstly, we prove that $u,v,h$ are linearly dependent in the case that $JH$ is nilpotent and $\deg_{z}v(x,y,z)=1$ and $\deg_{z}u\geq 2$. In the proof of the following theorem, we divide two cases according to the degree of $z$ in $u$. Case I: $\deg_{z}u=2$; Case II: $\deg_{z}u\geq 3$. In the first case, we divide two parts. We have three subcases in the second case.

Theorem 4.1.

Let $H=(u(x,y,z),v(x,y,z),h(x,y))$ be a polynomial map with $\deg_{z}v(x,y,z)=1$. Assume that $H(0)=0$ and $\deg_{z}u\geq 2$. If $JH$ is nilpotent, then $u,v,h$ are linearly dependent.

Proof.

Let $u=u_{d}z^{d}+\cdots+u_{1}z+u_{0}$, $v=v_{1}z+v_{0}$ with $u_{d}v_{1}\neq 0$ and $d\geq 2$. Since $JH$ is nilpotent, we have the following equations:

$$\left\{\begin{aligned} u_{x}+v_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.1)\\ u_{x}v_{y}-v_{x}u_{y}-h_{x}u_{z}-h_{y}v_{z}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.2)\\ v_{x}h_{y}u_{z}-h_{x}v_{y}u_{z}+h_{x}u_{y}v_{z}-u_{x}h_{y}v_{z}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.3)\\ \end{aligned}\right.$$

It follows from equation (4.1) that

$$u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{1y}z+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.4)$$

We always view that the polynomials are in ${\bf K}[x,y][z]$ with coefficients in ${\bf K}[x,y][z]$ in the following arguments. Thus, we have

$$u_{dx}=\cdots=u_{2x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.5)$$

and

$$u_{ix}+v_{iy}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.6)$$

for $i=1,~{}0$ by comparing the coefficients of $z^{j}$ of equation $(4.4)$ for $j=d,\ldots,1,0$.

It follows from equations (4.2) and (4.5) that
$(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-(v_{1x}z+v_{0x})(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0$                          (4.7)

Case I If $d=2$, then comparing the coefficients of $z^{3}$ and $z^{2}$ of equation (4.7), we have $v_{1x}u_{2y}=0$ and

$$u_{1x}v_{1y}-v_{1x}u_{1y}-v_{0x}u_{2y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.8)$$

Thus, we have $u_{2y}=0$ or $v_{1x}=0$.

(1) If $u_{2y}=0$, then it follows from equations (4.5) and (4.8) that $u_{2}\in{\bf K}^{*}$ and

$$u_{1x}v_{1y}-v_{1x}u_{1y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.9)$$

It follows from equation (4.6) that $u_{1x}=-v_{1y}$, so there exists $P\in{\bf K}[x,y]$ such that $u_{1}=P_{y}$, $v_{1}=-P_{x}$. It follows from equation (4.9) that $P_{xy}^{2}-P_{xx}P_{yy}=0$. Then it follows from Lemma 2.1 in [4] that

$$u_{1}=P_{y}=bf(ax+by)+c_{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.10)$$

and

$$v_{1}=-P_{x}=-af(ax+by)+c_{1}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.11)$$

for some $f(t)\in{\bf K}[t]$ and $f(0)=0$, $a,b\in{\bf K}^{*}$, $c_{1},c_{2}\in{\bf K}$. Then equation (4.7) has the following form:

$$(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-(v_{1x}z+v_{0x})(u_{1y}z+u_{0y})-h_{x}(2u_{2}z+u_{1})-h_{y}v_{1}=0~{}~{}~{}(4.12)$$

Thus, we have the following equations

$$u_{1x}v_{0y}+u_{0x}v_{1y}-v_{1x}u_{0y}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.13)$$

and

$$u_{0x}v_{0y}-v_{0x}u_{0y}-h_{x}u_{1}-h_{y}v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.14)$$

by comparing the coefficients of $z$ and $z^{0}$ of equation (4.12). It follows from equations (4.3) and (4.5) that

$$[(v_{1x}z+v_{0x})h_{y}-h_{x}(v_{1y}z+v_{0y})](2u_{2}z+u_{1})+[h_{x}(u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0~{}~{}~{}(4.15)$$

Comparing the coefficients of $z^{2},z,z^{0}$ of equation $(4.15)$, we have the following equations:

$$v_{1x}h_{y}-h_{x}v_{1y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.16)$$

and

$$(v_{0x}h_{y}-h_{x}v_{0y})2u_{2}+(v_{1x}h_{y}-h_{x}v_{1y})u_{1}+(h_{x}u_{1y}-h_{y}u_{1x})v_{1}=0~{}~{}~{}~{}~{}~{}(4.17)$$

and

$$(v_{0x}h_{y}-h_{x}v_{0y})u_{1}+(h_{x}u_{0y}-h_{y}u_{0x})v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.18)$$

It follows from equations (4.11) and (4.16) that $af^{\prime}\cdot(bh_{x}-ah_{y})=0$. Thus, we have $a=0$ or $f^{\prime}=0$ or $bh_{x}=ah_{y}$.

(i) If $a=0$, then

$$u_{1}=bf(by)+c_{2},~{}~{}~{}v_{1}=c_{1}\in{\bf K}^{*}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.19)$$

It follows from equation (4.13) that $2u_{2}h_{x}=-v_{0x}u_{1y}$. That is,

$$2u_{2}h_{x}=-b^{2}f^{\prime}(by)v_{0x}.$$

Integrating with respect to $x$ of two sides of the above equation, we have

$$h=-\frac{b^{2}}{2u_{2}}f^{\prime}(by)v_{0}+\frac{c(y)}{2u_{2}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.20)$$

for some $c(y)\in{\bf K}[y]$. Substituting equations (4.19) and (4.20) to equation (4.17), we have the following equation:

$$v_{0x}[-b^{3}f^{\prime\prime}(by)v_{0}+c^{\prime}(y)-\frac{b^{4}}{2u_{2}}v_{1}\cdot(f^{\prime}(by))^{2}]=0.$$

Thus, we have $v_{0x}=0$ or $f^{\prime\prime}(by)=0$ and $c^{\prime}(y)=\frac{b^{4}}{2u_{2}}v_{1}\cdot(f^{\prime}(by))^{2}$.

If $v_{0x}=0$, then it follows from equations (4.13) and (4.19) that $h_{x}=0$. It follows from equation (4.18) that $v_{1}h_{y}u_{0x}=0$. Thus, we have $h_{y}=0$ or $u_{0x}=0$. If $u_{0x}=0$, then it follows from equation (4.14) that $h_{y}=0$. If $h_{y}=0$, then $h=0$ because $h(0,0)=0$. Thus, $u,v,h$ are linearly dependent.

If $f^{\prime\prime}(by)=0$, then $f^{\prime}(by)\in{\bf K}$ and $c^{\prime}(y)=\frac{b^{4}}{2u_{2}}v_{1}(f^{\prime}(by))^{2}\in{\bf K}$. Let $l:=b^{2}f^{\prime}(by)$ and $c:=\frac{l^{2}}{(2u_{2})^{2}}v_{1}$. Then it follows from equation (4.19) that $v_{1}\in{\bf K}^{*}$ and $u_{1}=ly+c_{2}$ for some $c_{2}\in{\bf K}$. Since $h(0,0)=0$, so it follows from equation (4.20) that

$$h=-\frac{l}{2u_{2}}v_{0}+c\cdot y~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.21)$$

It follows from equations (4.14) and (4.21) that

$$v_{0x}u_{0y}-u_{0x}v_{0y}=\frac{l}{2u_{2}}u_{1}v_{0x}+\frac{l}{2u_{2}}v_{1}v_{0y}-cv_{1}~{}~{}~{}~{}~{}~{}~{}~{}(4.22)$$

It follows from equations (4.18) and (4.21) that

$$c\cdot u_{1}v_{0x}+v_{1}[-\frac{l}{2u_{2}}(v_{0x}u_{0y}-u_{0x}v_{0y})-c\cdot u_{0x}]=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.23)$$

Substituting equation (4.22) to equation (4.23), we have the following equation:

$$c\cdot u_{1}v_{0x}+v_{1}[-\frac{l^{2}}{(2u_{2})^{2}}u_{1}v_{0x}+\frac{lc}{2u_{2}}v_{1}-\frac{l^{2}}{(2u_{2})^{2}}v_{1}v_{0y}-c\cdot u_{0x}]=0.$$

Since $c=\frac{l^{2}}{(2u_{2})^{2}}v_{1}$, so the above equation has the following form:

$$\frac{lc}{2u_{2}}v_{1}-c(v_{0y}+u_{0x})=0.$$

Substituting equation (4.6)$(i=0)$ to the above equation, we have $\frac{lc}{2u_{2}}v_{1}=0$. That is, $lc=0$. Since $c=\frac{l^{2}}{(2u_{2})^{2}}v_{1}$, so we have $c=l=0$. It follows from equation (4.21) that $h=0$. Thus, $u,v,h$ are linearly dependent.

(ii) If $f^{\prime}=0$, then $f=0$ because $f(0)=0$. That is, $u_{1}=c_{2}$, $v_{1}=c_{1}\in{\bf K}^{*}$. It follows from equation (4.13) that $h_{x}=0$. It follows from equation (4.17) that $v_{0x}h_{y}=0$. Thus, we have $h_{y}=0$ or $v_{0x}=0$.
If $h_{y}=0$, then $h=0$ because $h(0,0)=0$. Thus, $u,v,h$ are linearly dependent.
If $v_{0x}=0$, then it follows from equation (4.18) that $u_{0x}h_{y}=0$. That is, $u_{0x}=0$ or $h_{y}=0$. If $u_{0x}=0$, then it follows from equation (4.14) that $h_{y}=0$. Thus, we have that $h_{y}=0$. It reduces to the above case.

(iii) If $bh_{x}=ah_{y}$, then we have

$$h_{y}=\frac{b}{a}h_{x}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.24)$$

because we can assume that $a\cdot f^{\prime}\neq 0$. Otherwise, it reduces to the former two cases. Let $\bar{x}=ax+by$, $\bar{y}=y$. Then we have that $h_{\bar{y}}=0$. That is, $h\in{\bf K}[ax+by]$. It follows from equations (4.10), (4.11), (4.17),(4.24) that

$$h_{x}(\frac{b}{a}v_{0x}-v_{0y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.25)$$

It follows from equations (4.18) and (4.24) that

$$u_{1}h_{x}(\frac{b}{a}v_{0x}-v_{0y})+v_{1}h_{x}(u_{0y}-\frac{b}{a}u_{0x})=0~{}~{}~{}~{}~{}~{}~{}(4.26)$$

It follows from equations (4.25) and (4.26) that $h_{x}=0$ or $bv_{0x}=av_{0y}$ and $bu_{0x}=au_{0y}$.
If $h_{x}=0$, then it follows from equation (4.24) that $h_{y}=0$. Thus, we have $h=0$ because $h(0,0)=0$. so $u,v,h$ are linearly dependent.
If $bv_{0x}=av_{0y}$ and $bu_{0x}=au_{0y}$, then $v_{0},u_{0}\in{\bf K}[ax+by]$. Thus, it follows from equations (4.10), (4.11) and (4.13) that $h_{x}=0$. It reduces to the former case.

(2) If $v_{1x}=0$, then it follows from equation (4.8) that

$$u_{1x}v_{1y}-v_{0x}u_{2y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.27)$$

It follows from equation (4.7) that

$$(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-v_{0x}(u_{2y}z^{2}+u_{1y}z+u_{0y})-h_{x}(2u_{2}z+u_{1})-h_{y}v_{1}=0~{}~{}~{}~{}(4.28)$$

Comparing the coefficients of $z^{2},~{}z,~{}z^{0}$ of equation (4.28), we have the following equations:

$$\left\{\begin{aligned} u_{1x}v_{1y}-v_{0x}u_{2y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.29)\\ u_{1x}v_{0y}+u_{0x}v_{1y}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}(4.30)\\ u_{0x}v_{0y}-v_{0x}u_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.31)\\ \end{aligned}\right.$$

It follows from equations (4.3) and (4.5) that

$$[v_{0x}h_{y}-(v_{1y}z+v_{0y})h_{x}](2u_{2}z+u_{1})+[h_{x}(u_{2y}z^{2}+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0$$

Comparing the coefficients of $z^{2},~{}z$ and $z^{0}$ of the above equation, we have the following equations:

$$\left\{\begin{aligned} h_{x}(v_{1}u_{2y}-2u_{2}v_{1y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.32)\\ -v_{1y}h_{x}u_{1}+2u_{2}(v_{0x}h_{y}-v_{0y}h_{x})+v_{1}(h_{x}u_{1y}-h_{y}u_{1x})=0~{}~{}~{}~{}(4.33)\\ u_{1}(v_{0x}h_{y}-v_{0y}h_{x})+v_{1}(h_{x}u_{0y}-h_{y}u_{0x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.34)\\ \end{aligned}\right.$$

It follows from equation (4.32) that $h_{x}=0$ or $v_{1}u_{2y}=2u_{2}v_{1y}$.

If $h_{x}=0$, then it follows from equation (4.33) that $h_{y}(2u_{2}v_{0x}-u_{1x}v_{1})=0$. Thus, we have that $h_{y}=0$ or $2u_{2}v_{0x}=v_{1}u_{1x}$. If $h_{y}=0$, then $h=0$ because $h(0,0)=0$. Thus, $u,v,h$ are linearly dependent. If $2u_{2}v_{0x}=v_{1}u_{1x}$, then it follows from equation (4.6) that

$$2u_{2}v_{0x}=-v_{1}v_{1y}\in{\bf K}[y]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.35)$$

Substituting equations (4.35) and (4.6) to equation (4.29), we have the following equation: $v_{1y}(2u_{2}v_{1y}-v_{1}u_{2y})=0$. Thus, we have $v_{1y}=0$ or $2u_{2}v_{1y}=v_{1}u_{2y}$.
If $v_{1y}=0$, then it follows from equation (4.29) that $v_{0x}u_{2y}=0$. Thus, we have $v_{0x}=0$ or $u_{2y}=0$. If $u_{2y}=0$, then it reduces to Case I(1). If $v_{0x}=0$, then it follows form equation (4.34) that $h_{y}u_{0x}=0$. Thus, we have $h_{y}=0$ or $u_{0x}=0$. If $u_{0x}=0$, then it follows from equation (4.31) that $h_{y}=0$. Therefore, we have $h=0$ because $h(0,0)=0$. So $u,v,h$ are linearly dependent.

If $2u_{2}v_{1y}=v_{1}u_{2y}$, then we have

$$\frac{u_{2y}}{u_{2}}=2\frac{v_{1y}}{v_{1}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.36)$$

Suppose $u_{2y}v_{1y}\neq 0$. Then we have

$$u_{2}=e^{\bar{c}(x)}v_{1}^{2}$$

by integrating of two sides of equation (4.36) with respect to $y$. where $\bar{c}(x)$ is a function of $x$. Since $u_{2},v_{1}\in{\bf K}[x,y]$, we have $e^{\bar{c}(x)}\in{\bf K}[x]$. That is, $u_{2}=c(x)v_{1}^{2}$ for $c(x)\in{\bf K}[x]$ and $c(x)\neq 0$. Then it follows from equations (4.29) and (4.6) that

$$v_{1y}(2c(x)v_{1}v_{0x}+v_{1y})=0.$$

That is,

$$2c(x)v_{1}v_{0x}=-v_{1y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.37)$$

If $v_{0x}\neq 0$, then we have that $v_{1y}=0$ by comparing the degree of $y$ of equation (4.37). Thus, we have $v_{0x}=0$. This is a contradiction. Therefore, we have $v_{1y}=v_{0x}=0$. It follows from equation (4.36) that $u_{2y}=0$, This is a contradiction! If $u_{2y}v_{1y}=0$, then it follows from equation (4.36) that $u_{2y}=v_{1y}=0$. which reduces to case I(1).

Case II If $d\geq 3$, then we have

$$u_{dy}v_{1x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.38)$$

and

$$v_{1x}u_{(d-1)y}+v_{0x}u_{dy}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.39)$$

by comparing the coefficients of $z^{d+1}$ and $z^{d}$ of equation (4.7). It follows from equations (4.38) and (4.39) that $v_{1x}=v_{0x}=0$ or $v_{1x}=0=u_{dy}$ or $u_{dy}=u_{(d-1)y}=0$.

(a) If $v_{1x}=v_{0x}=0$, then equation (4.7) has the following form:

$$(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.40)$$

If $d>3$, then $h_{x}=0$ by comparing the coefficient of $z^{d-1}$ of equation (4.40). Thus, it follows from equation (4.3) that $h_{y}(u_{1x}z+u_{0x})=0$. Therefore, we have $h_{y}=0$ or $u_{1x}=u_{0x}=0$. If $u_{1x}=u_{0x}=0$, then it follows from equation (4.40) that $h_{y}=0$. Thus, we have $h=0$ because $h(0,0)=0$. Therefore, $u,v,h$ are linearly dependent.
If $d=3$, then comparing the coefficients of $z^{2},z$ and $z^{0}$ of equation (4.40), we have the following equations:

$$\left\{\begin{aligned} u_{1x}v_{1y}-3u_{3}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.41)\\ u_{1x}v_{0y}-v_{1y}u_{0x}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}(4.42)\\ u_{0x}v_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.43)\\ \end{aligned}\right.$$

It follows from equations (4.3) and (4.5) that $-h_{x}(v_{1y}z+v_{0y})(3u_{3}z^{2}+2u_{2}z+u_{1})+[h_{x}(u_{3y}z^{3}+u_{2y}z^{2}+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0$                 (4.44)
Comparing the coefficients of $z^{3}$ of the above equation, we have $h_{x}(3v_{1y}u_{3}-u_{3y}v_{1})=0$. Thus, we have $h_{x}=0$ or $3u_{3}v_{1y}=v_{1}u_{3y}$.

(1) If $h_{x}=0$, then it follows from equation (4.44) that $h_{y}(u_{1x}z+u_{0x})=0$. That is, $h_{y}=0$ or $u_{1x}=u_{0x}=0$. If $u_{1x}=u_{0x}=0$, then it follows from equation (4.40) that $h_{y}=0$. Thus, we have $h=0$ because $h(0,0)=0$. Therefore, $u,v,h$ are linearly dependent.

(2) If $3u_{3}v_{1y}=v_{1}u_{3y}$, then

$$\frac{u_{3y}}{u_{3}}=3\frac{v_{1y}}{v_{1}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.45)$$

If $v_{1y}=0$, then $u_{3y}=0$. It follows from equation (4.41) that $h_{x}=0$. Then it follows from the arguments of Case II (1) that $u,v,h$ are linearly dependent. We can assume that $u_{3y}v_{1y}\neq 0$. Then we have that $u_{3}=e^{\bar{d}(x)}v_{1}^{3}$ by integrating the two sides of equation (4.45) with respect to $y$, where $\bar{d}(x)$ is a function of $x$. Since $u_{3},v_{1}\in{\bf K}[x,y]$, we have $e^{\bar{d}(x)}\in{\bf K}[x]$. That is,

$$u_{3}=d(x)v_{1}^{3}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.46)$$

with $d(x)\in{\bf K}[x]$, $d(x)\neq 0$. Substituting equations (4.6) and $(4.46)$ to equation (4.41), we have that

$$-3d(x)v_{1}^{3}h_{x}=v_{1y}^{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.47)$$

Then we have $v_{1y}=0$ by comparing the degree of $y$ of equation (4.47). It follows from equation (4.47) that $h_{x}=0$. Then it follows from the arguments of Case II(1) that $u,v,h$ are linearly dependent.

(b) If $v_{1x}=u_{dy}=0$, then it follows from equation (4.6) that $u_{d}\in{\bf K}^{*}$. then equation (4.7) has the following form: $(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-v_{0x}(u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0$                     (4.48)
Then we have the following equations:

$$-v_{0x}u_{iy}-(i+1)u_{i+1}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.49)$$

by comparing the coefficients of $z^{i}$ of equation (4.49) for $i=d-1,d-2,\ldots,3$. Comparing the coefficients of $z^{2},z$ and $z^{0}$ of equation (4.48), we have the following equations:

$$\left\{\begin{aligned} u_{1x}v_{1y}-v_{0x}u_{2y}-3u_{3}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.50)\\ u_{1x}v_{0y}+v_{1y}u_{0x}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}(4.51)\\ u_{0x}v_{0y}-v_{0x}u_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.52)\\ \end{aligned}\right.$$

It follows from equations (4.3) and (4.5) that
$[v_{0x}h_{y}-h_{x}(v_{1y}z+v_{0y})](du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})+[h_{x}(u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0$                       (4.53)
Then we have $h_{x}v_{1y}=0$ by comparing the coefficients of $z^{d}$ of equation (4.53). That is, $h_{x}=0$ or $v_{1y}=0$.

(3) If $h_{x}=0$, then equation (4.53) has the following form :

$$v_{0x}h_{y}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})-h_{y}(u_{1x}z+u_{0x})v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.54)$$

Comparing the coefficients of $z^{d-1}$ of equation (4.54), we have that $v_{0x}h_{y}=0$. That is, $v_{0x}=0$ or $h_{y}=0$. If $v_{0x}=0$, then it reduces to Case II(a). If $h_{y}=0$, then $h=0$ because $h(0,0)=0$. Thus, $u,v,h$ are linearly dependent.

(4) If $v_{1y}=0$, then comparing the coefficients of $z^{d-1}$ and $z^{0}$ of equation (4.53), we have

$$(v_{0x}h_{y}-h_{x}v_{0y})du_{d}+h_{x}u_{(d-1)y}v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.55)$$

and

$$(v_{0x}h_{y}-h_{x}v_{0y})u_{1}+(h_{x}u_{0y}-h_{y}u_{0x})v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.56)$$

If $d>3$, then it follows from equation (4.49)$(i=d-1)$ that

$$h_{x}=-\frac{1}{du_{d}}v_{0x}u_{(d-1)y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.57)$$

If $d=3$, then it follows from equation (4.50) that

$$h_{x}=-\frac{1}{3u_{3}}v_{0x}u_{2y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.57)$$

Substituting equation (4.57) to equation (4.55), we have

$$v_{0x}[h_{y}-\frac{v_{1}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}+\frac{1}{du_{d}}u_{(d-1)y}v_{0y}]=0$$

for $d\geq 3$. Thus, we have $v_{0x}=0$ or

$$h_{y}=\frac{v_{1}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}-\frac{1}{du_{d}}u_{(d-1)y}v_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.58)$$

If $v_{0x}=0$, then it reduces to Case II(a). Otherwise, substituting equations (4.57) and (4.58) to equation (4.52),we have that

$$u_{0x}v_{0y}-v_{0x}u_{0y}=-\frac{u_{1}}{du_{d}}v_{0x}u_{(d-1)y}+\frac{v_{1}^{2}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}-\frac{v_{1}}{du_{d}}u_{(d-1)y}v_{0y}~{}~{}~{}~{}~{}~{}~{}(4.59)$$

Substituting equations (4.57) and (4.58) to equation (4.56), we have that

$$\frac{u_{1}v_{1}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}v_{0x}-\frac{v_{1}^{2}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}u_{0x}+\frac{v_{1}}{du_{d}}u_{(d-1)y}(u_{0x}v_{0y}-v_{0x}u_{0y})=0~{}~{}~{}~{}~{}(4.60)$$

Then we have $\frac{v_{1}^{3}}{d^{3}u_{d}^{3}}u_{(d-1)y}^{3}=0$ by substituting equations (4.6) and (4.59) to equation (4.60). That is, $u_{(d-1)y}=0$. It follows from equation (4.49)$(i=d-1)$ that $h_{x}=0$. Then it reduces to Case II (3).

(c) If $u_{dy}=u_{(d-1)y}=0$, then it follows from equations (4.2) and (4.5) that $(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-(v_{1x}z+v_{0x})(u_{(d-2)y}z^{d-2}+\cdots+u_{1y}z+u_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0$                                      (4.61)
Comparing the coefficients of $z^{j}$ of equation (4.61) for $j=d-2,\ldots,3$, we have the following equations:

$$-v_{1x}u_{(j-1)y}-v_{0x}u_{jy}-(j+1)u_{j+1}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.62)$$

Comparing the coefficients of $z^{d-1},z^{2},z$ and $z^{0}$, we have the following equations:

$$\left\{\begin{aligned} -v_{1x}u_{(d-2)y}-du_{d}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.63)\\ u_{1x}v_{1y}-v_{1x}u_{1y}-v_{0x}u_{2y}-3u_{3}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.64)\\ u_{1x}v_{0y}+v_{1y}u_{0x}-v_{1x}u_{0y}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}(4.65)\\ u_{0x}v_{0y}-v_{0x}u_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.66)\\ \end{aligned}\right.$$

It follows from equations (4.3) and (4.5) that
$[(v_{1x}z+v_{0x})h_{y}-h_{x}(v_{1y}z+v_{0y})](du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})+[h_{x}(u_{(d-2)y}z^{d-2}+\cdots+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0$                              (4.67)
Comparing the coefficients of $z^{d}$ and $z^{d-1}$ of equation (4.67), we have the following equations:

$$\left\{\begin{aligned} du_{d}(v_{1x}h_{y}-h_{x}v_{1y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\ (d-1)u_{d-1}(v_{1x}h_{y}-h_{x}v_{1y})+du_{d}(v_{0x}h_{y}-h_{x}v_{0y})=0\\ \end{aligned}\right.$$

That is, $v_{1x}h_{y}-h_{x}v_{1y}=0$ and $v_{0x}h_{y}-h_{x}v_{0y}=0$. Then equation (4.67) has the following form :

$$h_{x}(u_{(d-2)y}z^{d-2}+\cdots+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.68)$$

Then we have $h_{x}u_{ky}=0$ by comparing the coefficients of $z^{k}$ of equation (4.68) for $2\leq k\leq d-2$. Thus, we have $h_{x}=0$ or $u_{(d-2)y}=\cdots=u_{2y}=0$. If $u_{(d-2)y}=\cdots=u_{2y}=0$, then it follows from equation (4.63) that $h_{x}=0$. If $h_{x}=0$, then it follows from equation (4.68) that $h_{y}=0$ or $u_{1x}=u_{0x}=0$. If $u_{1x}=u_{0x}=0$, then it follows from equation (4.63) and (4.62) that $v_{1x}=0$ or $u_{(d-2)y}=\cdots=u_{2y}=0$. If $v_{1x}=0$, then it reduces to Case II(b). If $u_{(d-2)y}=\cdots=u_{2y}=0$, then it follows from equation (4.64) that $u_{1y}=0$. It follows from equation (4.65) that $u_{0y}=0$. Then it follows from equation (4.66) that $h_{y}=0$. Thus, we have $h=0$ because $h(0,0)=0$. Therefore, $u,v,h$ are linearly dependent. ∎

Corollary 4.2.

Let $H=(u(x,y,z),v(x,y,z),h(x,y))$ be a polynomial map with $\deg_{z}v(x,y,z)\leq 1$. Assume that $H(0)=0$ and the components of $H$ are linearly independent over ${\bf K}$. If $JH$ is nilpotent, then
$(1)$ $H$ has the form of Theorem 2.2 in the case $\deg_{z}u=0$ and $(\deg_{y}u,\deg_{y}h)\leq 3$.
$(2)$ $H$ has the form of Proposition 3.1 in the case $\deg_{z}v=0$ and $(\deg_{x}v,\deg_{x}h)\leq 3$.
$(3)$ there exists $T\in\operatorname{GL}_{3}({\bf K})$ such that $T^{-1}HT$ has the form of Theorem 2.2 in the case $\deg_{z}u\cdot\deg_{z}v\geq 1$ and $(\deg(u(x,y,0)-\lambda v(x,y,0)),\deg h(x,y))\leq 3$ for some $\lambda\in{\bf K}$.

Proof.

Since $u,v,h$ are linearly independent, so it follows from Theorem 4.1 that $\deg_{z}u\leq 1$. Thus, we have $\deg_{z}u\leq 1$ and $\deg_{z}v\leq 1$. Then the conclusions (1) and (2) follow from Theorem 2.2 and Proposition 3.1 respectively. The conclusion of (3) follows from the proof of Theorem 3.10 in [12] and Theorem 2.2. ∎